How many grams of salt is formed in the reaction between sulfuric acid and 40.5 g of aluminum?

univerkov | January 17, 2021 | education

Given:
m (Al) = 40.5 g
Find: m (Al2 (SO4) 3)
Decision:
2Al + 3H2SO4 = Al2 (SO4) 3 + 3H2
n (Al) = m / M = 40.5 g / 27 g / mol = 1.5 mol
n (Al): n (Al2 (SO4) 3) = 2: 1
n (Al2 (SO4) 3) = 1.5 mol / 2 = 0.75 mol
m (Al2 (SO4) 3) = n * M = 0.75 mol * 342 g / mol = 256.5 g
Answer: 256.5 g

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